The graphs of the functions $f(x)=\sin(x)$ and $g(x)=\dfrac{1}{2}$ intersect at $2$ points on the interval $0 < x < \pi$. What is the area of the region bound by the graphs of $f(x)$ and $g(x)$ between those points of intersection ? Choose 1 answer: Choose 1 answer: (Choice A) A $2-\dfrac{\pi}{2}$ (Choice B) B $\dfrac{\pi}{2}$ (Choice C) C $\sqrt{3}-\dfrac{\pi}{3}$ (Choice D) D $\dfrac{\pi}{3}$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${1}$ ${2}$ ${3}$ ${1}$ $f$ $g$ $y$ $x$ From the graph, it appears that $f(x)\ge g(x)$ between the points where the graphs intersect. From this we are looking to evaluate: $ \int_{a}^{b}\left( f(x)-g(x) \right)\,dx$ where $a$ and $b$ are the $x$ -coordinates of the points of intersection. Finding the $x$ -coordinates of the intersection points We can find the $x$ -coordinate of each point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{aligned} f(x)&=g(x) \\\\ \sin(x)&=\dfrac{1}{2} \end{aligned}$ The graphs intersect when $x=\dfrac{\pi}{6}\pm 2k\pi$ and when $x=\dfrac{5\pi}{6} \pm 2k\pi$, where $k$ is any integer. Since we want values of $x$ on the interval $0 < x < \pi$, we will use $x=\dfrac{\pi}{6}$ and $x= \dfrac{5\pi}{6}$. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $ \int_{\pi/6}^{5\pi/6}\left(\sin(x)-\dfrac{1}{2}\right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{\pi/6}^{5\pi/6} \left( \sin(x)- \dfrac{1}{2} \right) \,dx \\\\ &=-\cos(x)-\dfrac{x}{2}~ \Bigg|_{\pi/6}^{5\pi/6} \\\\ &= \left( \dfrac{\sqrt{3}}{2} -\dfrac{5\pi}{12}\right) - \left( -\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{12} \right) \\\\ &= \sqrt{3}-\dfrac{\pi}{3} \end{aligned}$ Answer The area is $\sqrt{3}-\dfrac{\pi}{3}$ square units.